TS EAMCET 9 May 2024 Shift 1 Solved Paper

In an intrinsic semiconductor, the concentration of electrons $(n_i)$ and holes $(p_i)$ are equal and given by:$n_i = p_i = 6 \times 10^{15} \text{m}^{-3} .$When the semiconductor is doped with an impurity, the electron concentration increases to $4 \times 10^{22} \text{m}^{-3}$, which is denoted by $n$. The product of the electron concentration ( $n$ ) and the hole concentration ( $p$ ) in a semiconductor at thermal equilibrium is given by:$n \cdot p = n_i^2 .$To find the hole concentration $(p)$ in the doped semiconductor, use the above relationship:Calculate $n_i^2$ :$n_i^2 = (6 \times 10^{15} \text{m}^{-3})^2 = 36 \times 10^{30} \text{m}^{-6} .$Using the relation $n \cdot p = n_i^2$, solve for $p$ :$p = \; \frac{n_i^2}{n} = \; \frac{36 \times 10^{30} \text{m}^{-6}}{4 \times 10^{22} \text{m}^{-3}} .$Calculate $p = \; \frac{36}{4} \times 10^{30-22} \text{m}^{-3} = 9 \times 10^8 \text{m}^{-3} .$ :$9 \times 10^8 \text{m}^{-3}$Therefore, the concentration of holes in the doped semiconductor is $\; 18 \times 10^{-8} \text{m}^{-3}$, which corresponds to Option C.